A particle performing linear S.H.M. has period 8 second. At time $t=0$, it is in the mean position. The ratio of the distances travelled by the particle in the $1^{\text {st }}$ and $2^{\text {nd }}$ second is $\left(\cos 45^{\circ}=1 / \sqrt{2}\right)$

The motion is described by $x = A \sin(\omega t)$ with period $T = 8\,\text{s}$, giving angular frequency $\omega = \frac{2\pi}{T} = \frac{\pi}{4}\,\text{rad/s}$.

First second: The displacement at $t = 1\,\text{s}$ is $x_1 = A \sin\left(\frac{\pi}{4}\right) = \frac{A}{\sqrt{2}}$. Since motion starts from the mean position, the distance traveled is $d_1 = \frac{A}{\sqrt{2}}$.

Second second: The displacement reaches $x_2 = A \sin\left(\frac{\pi}{2}\right) = A$ at $t = 2\,\text{s}$. The distance covered from $t = 1\,\text{s}$ to $t = 2\,\text{s}$ is $d_2 = A - \frac{A}{\sqrt{2}} = A\left(1 - \frac{1}{\sqrt{2}}\right)$.

The ratio is $d_1 : d_2 = \frac{A}{\sqrt{2}} : A\left(1 - \frac{1}{\sqrt{2}}\right) = 1 : (\sqrt{2} - 1)$.